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Review of Classifying and Balancing Chemical Equations

Review of Classifying and Balancing Chemical Equations. Lets take a moment… sit back… relax… and review some previously learned concepts…. 5 main types of chemical reactions:. 1. Formation/Synthesis. 2 . Decomposition/De-formation. 3. Single Replacement. 4 . Double Replacement. 5.

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Review of Classifying and Balancing Chemical Equations

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  1. Review of Classifying and Balancing Chemical Equations • Lets take a moment… sit back… relax… and review some previously learned concepts…

  2. 5 main types of chemical reactions: 1. Formation/Synthesis 2. Decomposition/De-formation 3. Single Replacement 4. Double Replacement 5. Hydrocarbon Combustion 6. Other Combustions

  3. 1. Formation/Synthesis Reactions Reaction in which 2 or more pure elements combine to form a new compound. General Equation: A + B  AB Element + Element Compound Example: Na + Cl  NaCl (Sodium Chloride) SALT (Sodium) (Chlorine)

  4. 2. Decomposition/De-Formation Reaction in which a compound breaks down into the base elements that formed it. General Equation: AB  A + B Compound Element + Element Example: H2O(l)  H2(g) + O2(g) (Water) (Hydrogen) (Oxygen)

  5. 3. Single Replacement Reaction in which an element and a compound react. The “partners” are traded and the elements that make up the compound and lone element are substituted. General Equation: AB +C AC + B Compound + Element Compound + Element Example: NaCl(s) +Li(s) LiCl(s)+ Na(s) (Salt) (Soduim) (Lithium Chloride) (Lithium)

  6. 4. Double Replacement Reaction in which two compounds react and “switch partners”. The elements that make up each compound are swapped to form two new compounds. General Equation: AB +CD AD + CB Compound + Compound Compound + Compound Example: NaCl(s) +LiBr(s) LiCl(s)+ NaBr(s) (Salt) (Lithium Bromide) (Sodium Bromide) (Lithium Chloride)

  7. 5. Hydrocarbon Combustion Reaction in which a compound containing Hydrogen and Carbon are burned to produce H2O(g) and CO2(g). General Equation: CXHX +O2(g) H2O(g)+ CO2(g) CxHx+ Oxygen Water+ Carbon Dioxide Example: CH4(g) +O2(g) H2O(g)+ CO2(g) (Methane) (Oxygen) (Carbon Dioxide) (Water Vapour)

  8. 6. Other Combustions Reaction in which a compound OR element is burned forming its most common oxide. General Equation: X +O2(g) XO(g) X + Oxygen XO Example: X+O2(g) XO

  9. Steps to Balance a Chemical Equation • Write out the correct chemical formulae for the products and the reactants. Be sure to include all states of matter! • Balance the atoms or ion present in the greatest number. You may do this by finding the lowest common multiple of the two. • Continue to systematically balance the rest of the atoms or ions. • Check the final equation. Make sure every type of atom or ion balances.

  10. The coefficients that are placed in front of the compounds and elements represent the mole-to-mole ratio of the reactants and products. Ex. 2NaCl(aq) + MgSO4(aq)→ Na2SO4(aq) + MgCl2(s) • 2 moles of sodium chloride reacts with one mole of magnesium sulfate to produce one mole of sodium sulfate and one mole of magnesium chloride • When you are making quantitative calculations, please use the mole-to-mole ratio in your calculations :

  11. RXN Limitations • RXNS don’t communicate temp. and pressure • RXNS don’t communicate progress or process • RXNS don’t communicate measurable quantities of substances (can’t really weigh or calculate moles, atoms, molecules, ions, etc. without a calculation first)

  12. RXN Assumptions • RXNS are spontaneous • RXNS are fast • RXNS are quantitative (more than 99% complete) • RXNS are stoichiometric (simple, whole number ratio of chemical amounts of reactants and products

  13. Net Ionic Equations • Net ionic equations are useful in that they show only those chemical species participating in a chemical reaction • The key to being able to write net ionic equations is the ability to recognize monoatomic and polyatomic ions, and the solubility rules.

  14. Net Ionic Equations Step 1: Pb(NO3)2(aq) + 2HCl(aq) PbCl2(s) + 2 HNO3(aq) Step 2: Dissociate (split up) and soluble ionic compounds AND ionize (split up) all Strong Acids. + Pb2+(aq) 2 H3O+(aq) 2 Cl-(aq) + 2NO3-(aq) + 2NO3-(aq) +  2 H3O+(aq) + PbCl2(s)

  15. Net Ionic Equations PbCl2(s) 2 H3O+(aq) Pb2+(aq) 2 H3O+(aq) 2 Cl-(aq) + 2NO3-(aq) + 2NO3-(aq) + +  + Total Ionic Equation Step 3: Cancel out the spectator ions. (Ions that are the same on both sides) Step 4: Write down whatever is left over after canceling. Simplify balancing if necessary PbCl2(s) Pb2+(aq) 2 Cl-(aq) +  Net Ionic Equation

  16. BaCl2 + Na2SO4 2  NaCl BaSO4 + (aq) (aq) (aq) (s) Complete Balanced Chemical Equation What type of reaction? ****Have to use solubility table to determine the state (aq) or (s)**** Double Replacement Step 1: DONE

  17. BaCl2 + Na2SO4 2  NaCl BaSO4 + (aq) (aq) (aq) (s) Complete Balanced Chemical Equation Ba2+(aq) 2Cl-(aq) 2Na+(aq) SO42-(aq) 2Na+(aq) 2Cl-(aq) BaSO4(s) Step 2: Dissociate soluble ionic and ionize strong acids.

  18. 2Na+(aq) Ba2+(aq) + 2Cl-(aq) + 2Na+(aq)  + SO4+(aq) + 2Cl-(aq) + BaSO4(s) Step 2: DONE Step 3: Cancel out the spectator ions. (Ions that are the same on both sides) DONE Step 4: Write down whatever is left over after canceling. Simplify balancing if necessary. DONE SO4+(aq) BaSO4(s) Ba2+(aq) + 

  19. Gravimetric Stoichiometry: The process of calculating the masses involved in a chemical reaction using the balancing numbers and (Need/Got). Steps for Starting stoichiometry: • Balanced Chemical Equation. • Write information…mols….masse under the chemical/element that it relates to. • ***Convert information given in question into mols.*** • Find the chemical species the question wants to know about and put a N for (need to find) above it. • Put a G for (got information about this) over the species you have information about from the question.

  20. Step 1: Write a Balanced Chemical Equation for the reaction. n = m M 1 Fe2O3(s) 2 CO(g) +  Fe(s) 3 + 3 CO2(g) n = 100.0 g 55.85 g/mol m = ? m = 100.0g n =1.79 mol n = ? n =1.79 mol Step 2: Write the information from the question under the correct element/compound Convert information to mols. Step 3:

  21. Step 4: Put an N over the species you NEED to find out about. n = m M N G 1 Fe2O3(s) 2 CO(g) +  Fe(s) 3 + 3 CO2(g) n = 100.0 g 55.85 g/mol m = ? m = 100.0g n =1.79 mol n =1.79 mol Step 5: Put a G over the compound you know information about.

  22. Hydrogen gas and Oxygen gas react to produce water. How much hydrogen gas would be needed to produce 12.56 g of water? Step 1: Write a Balanced Chemical Equation for the reaction. N G 2 H2(g) + 1 O2(g) 2 H20(l) m = ? m = 12.56 g n = 0.697 mol Step 2: Write the information from the question under the correct element/compound Convert information to mols. Step 3: Step 4: Put an N over the species you NEED to find out about. Step 5: Put a G over the compound you know information about.

  23. Gravimetric Stoichiometry: The process of calculating the masses involved in a chemical reaction using the balancing numbers and (Need/Got). Steps for FINISHING stoichiometry: 6. Convert mols of got (G) to mols of need (N) by multiple by the conversion factor (N/G). 7. Convert the mols of need (N) into a mass by multiplying by the molar mass (M) of the needed compound.

  24. Step 6: Convert mols of got (G) to mols of need (N) by multiple by the conversion factor (N/G). N G Fe2O3(s) 1 2 CO(g) +  Fe(s) + 3 3 CO2(g) m = 100.0g m = ? n =0.895 mol n =1.79 mol 1 2 N G ( ) nFE x (1/2) = nFe2O3 (1.79mol) X (1/2) = nFe2O3 0.895 mol = nFe2O3

  25. Step 7: Convert mols of need (N) into mass of Need (N) by multiplying by N’s Molar Mass. n = m M N G Fe2O3(s) 1 2 CO(g) +  Fe(s) + 3 3 CO2(g) m = 142.93 g m = ? m = 100.0g m = nM m = (0.895 mol)( ) n =0.895 mol n =0.8953 mol 159.70 g/mol 1 2 ( ) 2 x 55.85 g/mol MFe2O3(s) = 3 x 16.00 g/mol 159.70 g/mol m = 142.93 g

  26. Hydrogen gas and Oxygen gas react to produce water. How much hydrogen gas would be needed to produce 12.56 g of water? N G 2 H2(g) + 1 O2(g) 2 H20(g) m = 1.4079 g m = ? m = 12.56 g n = 0.697 mol Step 6: Convert mols of got (G) to mols of need (N) by multiple by the conversion factor (N/G). Step 7: Convert the mols of need (N) into a mass by multiplying by the molar mass (M) of the needed compound.

  27. Excess and Limiting Reagents Excess and limiting reagents refer to the reactant that will run out first and stop more product from forming. Limiting Reagent Excess Reagent + 24 Crackers 7 Pieces of Cheese = Which will run out first?

  28. Excess and Limiting Reagents 12.4 g of Hydrogen gas reacts with 12.4 g of Oxygen gas to form liquid water. Determine the excess and limiting reagent. Calculate the maximum amount of water that could be produced by reacting these two gases together. G N H2(g) + O2(g)  H2O(g) 2 1 2 m = 12.4 g m = 12.4 g m = ? n = m M n = m M n = 6.138….mol n = 0.3875 mol ***From here on it’s like a double stoichiometry to see which can make more H2O(l)***

  29. Excess and Limiting Reagents 12.4 g of Hydrogen gas reacts with 12.4 g of Oxygen gas to form liquid water. Determine the excess and limiting reagent. Calculate the maximum amount of water that could be produced by reacting these two gases together. G N H2(g) + O2(g)  H2O(g) 2 1 2 m = ? m = 12.4 g m = 12.4 g n = 6.138….mol n = 0.3875 mol 2 2 N G ( ) H2 n = 6.138…

  30. Excess and Limiting Reagents 12.4 g of Hydrogen gas reacts with 12.4 g of Oxygen gas to form liquid water. Determine the excess and limiting reagent. Calculate the maximum amount of water that could be produced by reacting these two gases together. Limiting Excess N G H2(g) + O2(g)  H2O(g) 2 1 2 m = ? m = 12.4 g m = 12.4 g O2 n = 0.775 mol n = 6.138….mol n = 0.3875 mol 2 1 N G ( ) H2 n = 6.138…

  31. Percent Yield In every reaction you can use stoichiometry to calculate the theoretical amount of product that could be made. (Maximum or Total). When you actually do an experiment, the actual amount that you are able to make is called the actual amount. Actual Amount X 100% % Yield = Theoretical Amount

  32. Stoichiometry ALWAYS involves mols and a balanced chemical equation. H2(g) + O2(g)  H2O(g) 2 1 2 What information can get us n? What formula’s do we know that have n in them?

  33. H2(g) + O2(g)  H2O(g) 2 1 2 ALREADY DID THIS. Called Gravimetric Stoichiometry. n = m M Use this when we turn a mass into mols and do stoichiometry. SATP STP 24.8 mol/L 22.4 mol/L ***Because we can calculate mols we can then do stoichiometry just like before.*** n = 1 mol V = 24.8 L n = 1 mol V = 22.4 L

  34. 1 5 3 4 C3H8(g) + O2(g) CO2(g) + H2O(g) 6 4 10 Even Number

  35. N G 1 5 3 4 C3H8(g) + O2(g) CO2(g) + H2O(g) m = 275g V = 698 L V = ? 22.4 L 1 mol n = m M n = 31.2 mol nO2(g) @ STP = N G 5 1 ( ) n = 275g 44.11 g/mol = X 31.2 mol n = 6.23 mol X = 698 L

  36. (YOU TRY) G N 2 2 1 2 Na(s) + H2O(l)  H2(g) + NaOH(s) m= 37.1 g m = ? V = 20.0 L 24.8 L 1 mol m= nM m= (1.61mol)(22.99g/mol) m= 37.1 g nH2(g) @ SATP = n = m M = 20.0 L X 2 1 ( ) n = 0.806…mol n = 1.61…mol X = 0.80645 mol

  37. N 1 2 NH3(g) N2(g)  3 + H2(g) V = ? P = 450 kPa T = 333.15 K T = 80°C

  38. N G 1 2 NH3(g) N2(g)  3 + H2(g) V = ? m = 7.5 Kg m = 7500 g P = 450 kPa T = 333.15 K

  39. N G 1 2 NH3(g) N2(g)  3 + H2(g) V = ? m = 7500 g P = 450 kPa n = 3700 mol n = m M T = 333.15 K R = 8.314 n = 7500g 2.02 g/mol 2466.6.. mol 2 3 N G ( ) n = Have to calculate mols to do stoich.

  40. V = ? PV = nRT P = 450 kPa T = 333.15 K R = 8.314 2466.6.. mol V = (2466.6 mol)(8.314)(333.15) (450 kPa) n = V = 16000 L of NH3 V = 16 KL of NH3

  41. Stoichiometry ALWAYS involves mols and a balanced chemical equation. Solution Stoichiometry: Using the principles that govern concentrations to solve for n (mols). Once you have moles the stoichiometry is the same. C = n V ***Once you get mols…the rest is the same.***

  42. H2SO4(aq) +  KOH(aq) H2O(aq) + K2SO4(aq)

  43. 1 H2SO4(aq) + 2 1  2 KOH(aq) H2O(aq) + K2SO4(aq)

  44. N G 1 H2SO4(aq) + 2 1  2 KOH(aq) H2O(aq) + K2SO4(aq) V = 0.010 L V = 0.0159 L C = ? C = n V C = 0.150 mol/L n = 0.00119 n = CV n = (0.150 mol/L)(0.0159 L) 1 2 N G ( ) n = 0.00119 n = 0.00239 mol

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